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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Matrices

NDA 2018 · Mathematics · Matrices

If A=[ array ll 1 & 2 2 & 3 array ] and A^2-k A-I_2=O , where I_2...

If A=\left[1223\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] andA^2-k A-I_2=O,where, whereI_2 is the 2×22 \times 2 identity matrix, then what is the value of k ?

Options

  1. A4
  2. B-4
  3. C8
  4. D-8

Correct answer

A. 4

Step-by-step solution

A=[1223]A2=[1223][1223]=[1+42+62+64+9]=[58813]A2−kA−I2=[58813]−[k2k2k3k]−[1001]=[5−k−18−2k−08−2k−013−3k−1]=[4−k8−2k8−2k12−3k]\begin{aligned} & A=\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \\ & A^2=\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} 1+4 & 2+6 \\ 2+6 & 4+9 \end{array}\right]=\left[\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right] \\ & \mathrm{A}^2-\mathrm{kA}-\mathrm{I}_2=\left[\begin{array}{cc} 5 & 8 \\ 8 & 13 \end{array}\right]-\left[\begin{array}{cc} \mathrm{k} & 2 \mathrm{k} \\ 2 \mathrm{k} & 3 \mathrm{k} \end{array}\right]-\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 5-\mathrm{k}-1 & 8-2 \mathrm{k}-0 \\ 8-2 \mathrm{k}-0 & 13-3 \mathrm{k}-1 \end{array}\right] \\ & =\left[\begin{array}{cc} 4-\mathrm{k} & 8-2 \mathrm{k} \\ 8-2 \mathrm{k} & 12-3 \mathrm{k} \end{array}\right] \end{aligned} Given, A2−kA−I2=0∴4−k=0⇒k=4\begin{aligned} & \text{ Given, } \mathrm{A}^2-\mathrm{kA}-\mathrm{I}_2=0 \\ & \therefore 4-\mathrm{k}=0 \Rightarrow \mathrm{k}=4 \end{aligned}$
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